ta có:
\(3^{n+2}-2^{n+2}+3^n-2^n\\ =\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\\ =3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\\ =3^n\left(9+1\right)-2^n\left(4+1\right)\\ =3^n.10-2^n.5\\ =3^n.10-2^{n-1}.2.5\\ =3^n.10-2^{n-1}.10\\ =\left(3^n-2^{n-1}\right).10⋮10\\ \Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\left(đpcm\right)\)